Ok guys I need a little help on L' Hopital's question? calculus question.?

Question

show step by step how to find the limit using L' hopitals rule limit x-->0 (x2^x)/ (2^x-1)

P.S I have done my homework but this question for test review and I am getting wrong answer even when i used LN to find the answer.
Thank you for your help.

Answer 1

[x→0]lim (x2^x)/(2^x - 1)

This is indeed a 0/0 form, so differentiate both numerator and denominator. We have:

[x→0]lim (2^x + x2^x ln 2)/(2^x ln 2)

Which may now be evaluated directly:

(2^0 + 0*2^0 * ln 2)/(2^0 * ln 2)
1/ln 2

And we are done.

Answer 2

The derivative of x(2^x) is (x^2)(2^x)ln(2) + x(2^x); the derivative of 2^x - 1 is x(2^x)ln(2). The quotient of the derivatives is x + 1/ln(2) and the limit is 1/ln(2).